Problem 62
Question
Let the line \(\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}\) lies in the plane \(x+\) \(3 y-\alpha z+\beta=0 .\) Then \((\alpha, \beta)\) equals (A) \((6,-17)\) (B) \((-6,7)\) (C) \((5,-15)\) (D) \((-5,15)\)
Step-by-Step Solution
Verified Answer
The answer is (B) (-6, 7).
1Step 1: Write Direction Ratios of the Line
The given line equation is \(\frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}\). From this form, we can deduce that the direction ratios of the line are \((3, -5, 2)\).
2Step 2: Identify a Point on the Line
To find a point on the line, set the common ratio equal to a parameter, say \(t\). Then, \(x = 3t + 2\), \(y = -5t + 1\), \(z = 2t - 2\). For \(t = 0\), the point on the line simplifies to \((2, 1, -2)\).
3Step 3: Confirm Point Lies on the Plane
Substitute the point \((2, 1, -2)\) into the plane equation \(x + 3y - \alpha z + \beta = 0\): \[2 + 3(1) - \alpha(-2) + \beta = 0\] \[2 + 3 + 2\alpha + \beta = 0\] \[5 + 2\alpha + \beta = 0 .\]
4Step 4: Establish Relationship Using Normal to the Plane
The normal to the plane has the direction ratios \((1, 3, -\alpha)\). For the line to lie in the plane, the direction ratios of the line \((3, -5, 2)\) must be perpendicular to the plane's normal. Thus, \[3 \times 1 + (-5) \times 3 + 2 \times (-\alpha) = 0\] \[3 - 15 - 2\alpha = 0\]\[-2\alpha = 12\] \[\alpha = -6\].
5Step 5: Determine \(\beta\) Using the Plane Equation
Substitute \(\alpha = -6\) back into the equation found in Step 3: \[5 + 2(-6) + \beta = 0\] \[5 - 12 + \beta = 0\] \[\beta = 7\].
6Step 6: Verify Solution with Given Options
The calculated pair \((\alpha, \beta) = (-6, 7)\) matches option (B).
Key Concepts
Direction RatiosPlane and Line IntersectionPerpendicular Vectors
Direction Ratios
Direction ratios are an essential concept in geometry, especially when dealing with lines in three-dimensional space. They help define the orientation of a line. In this exercise, the line is given by the equation \(\frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}\). From this equation, we derive the direction ratios of the line as \((3, -5, 2)\).
Direction ratios are the coefficients that show the difference in each dimension due to a unit movement along the line. They provide a sense of direction without giving magnitude.
Direction ratios are the coefficients that show the difference in each dimension due to a unit movement along the line. They provide a sense of direction without giving magnitude.
- In our case, "3" tells us how much change occurs in the x-direction for each step along the line.
- Similarly, "-5" indicates the change in y-direction, and "2" represents the change in z-direction.
Plane and Line Intersection
Understanding the intersection between a plane and a line requires knowing both the line and the plane's equations. The intersection tells us where these two geometric entities meet, if at all. In our exercise, the line specified must be within the plane, not just intersect it.
The plane is represented by the equation \(x + 3y - \alpha z + \beta = 0\). A point on the line (found by substituting values like \(t = 0\) into the parametric form of the line) is \((2, 1, -2)\). By inserting this point into the plane equation, we derive the relationship \(5 + 2\alpha + \beta = 0\).
For a line to lie within a plane, every point on the line must satisfy the plane's equation, and its direction must be aligned perpendicular to the plane's normal vector. This ensures there is no intersection angle other than zero, meaning they coincide rather than merely intersect in a single point.
The plane is represented by the equation \(x + 3y - \alpha z + \beta = 0\). A point on the line (found by substituting values like \(t = 0\) into the parametric form of the line) is \((2, 1, -2)\). By inserting this point into the plane equation, we derive the relationship \(5 + 2\alpha + \beta = 0\).
For a line to lie within a plane, every point on the line must satisfy the plane's equation, and its direction must be aligned perpendicular to the plane's normal vector. This ensures there is no intersection angle other than zero, meaning they coincide rather than merely intersect in a single point.
Perpendicular Vectors
Whenever vectors are perpendicular, their dot product equals zero. This is a critical concept when dealing with planes and lines in geometry.
In our exercise, the normal to the plane has the direction ratios \((1, 3, -\alpha)\). For the line to lie completely in the plane, it must be perpendicular to the plane's normal.
By using the line's direction ratios \((3, -5, 2)\) and computing their dot product with the plane's normal, we get: \[3 \times 1 + (-5) \times 3 + 2 \times (-\alpha) = 0\]
In our exercise, the normal to the plane has the direction ratios \((1, 3, -\alpha)\). For the line to lie completely in the plane, it must be perpendicular to the plane's normal.
By using the line's direction ratios \((3, -5, 2)\) and computing their dot product with the plane's normal, we get: \[3 \times 1 + (-5) \times 3 + 2 \times (-\alpha) = 0\]
- The term \(3 \times 1\) represents the traditional dot product calculation for the x-coordinates.
- Similarly, \(-5 \times 3\) represents the y-components in the dot product.
Other exercises in this chapter
Problem 60
The line passing through the points \((5,1, a)\) and \((3, b,\), 1) crosses the \(y z\)-plane at the point \(\left(0, \frac{17}{2}, \frac{-13}{2}\right)\) then
View solution Problem 61
If the straight lines \(\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{3}\) and \(\frac{x-2}{3}=\frac{y-3}{k}=\frac{z-1}{2}\) intersect at a point, then the integer \(k
View solution Problem 63
A line \(\mathrm{AB}\) in 3 -dimensional space makes angles \(45^{\circ}\) and \(120^{\circ}\) with the positive \(x\)-axis and the positive \(y\)-axis respecti
View solution Problem 64
If the angle between the line \(x=\frac{y-1}{2}=\frac{z-3}{\lambda}\) and the plane \(x+2 y+3 z=\) is \(\cos ^{-4}\left(\sqrt{\frac{5}{14}}\right)\), then \(\la
View solution